Equilibrium with Multiple Objects: Study with Video Lessons, Practice Problems & Examples

In a seesaw problem, achieving rotational equilibrium involves balancing torques. The torque produced by a force is calculated as the product of the force and the distance from the fulcrum. For two children with masses m₁ and m₂ sitting at distances r₁ and r₂ from the fulcrum, the equation m₁r₁ = m₂r₂ is used. This relationship shows that the heavier child must sit closer to the fulcrum to maintain balance, illustrating the principle of torque in rotational dynamics.

Position of second kid on seesaw

Hey guys, so in this example, we're going to solve a problem where we have 2 kids playing on a seesaw, which is a classic problem in rotational equilibrium. And in this particular one, we have a kid all the way to the left, and we want to know how far we should place the kid on the right so that the system balances. Alright? So it says here the seesaw is 4 meters long. So I'm going to write l=4 meters. The mass of the seesaw, I'm going to call this big M, is 50. And the seesaw has uniform mass distribution. That means that the mg of the seesaw is in the middle, and it's pivoted on a fulcrum at its middle. So the fulcrum is at the middle, and the mg is at the middle as well. This means you have an mg down here, and you also have a normal force pushing this thing up at the fulcrum. Okay. The 2 kids sit at opposite sides. So one on the left, one on the right. The kid on the left has a mass of 30, and the kid on the right has a mass of 40. Okay. So first of all, you might imagine that if this is a perfectly symmetric system, you don't have to draw this, but if you have the bar held at the middle, and then you got the 30 here and the 40 here, this will tilt this way because the heavier kid, they have the same distance but the heavier kid has a bigger force pushing down so it tilts. So the solution then is to move the heavier kid closer to the middle. Okay. That's what we want to figure out is how far from the fulcrum should the kid be. Alright. So what we're doing here is we're looking for a situation where we have no rotation. So the sum of all torque will equal 0. And there are only 2 torques happening here. I have m1g at the left tip, which will cause a torque 1 over here, and then I have an m2g on the right side somewhere in the middle. That's going to cause a torque 2. Note that the torque of mg is 0. mg does not produce a torque, and that's because mg acts on the axis of rotation. Normal also doesn't produce a torque for the same reason. They both act on the axis of rotation. Forces that apply on the axis of rotation cause no torque. Since we want torques to cancel, we're just going to say that torque 1 has to equal torque 2. They're in opposite directions. So as long as they have the same magnitude, they will cancel. So I can write torque 1 equals torque 2. And once we expand this equation, that's where we'll be able to find our target variable. Alright? So torque 1 is due to m1gr11sin⁡θ1 and torque 2 is due to m2gr22sin⁡θ2. Just using the torque equation there, Notice the gravity is canceled because I have gravity on all terms. And what I'm looking for is r2. Alright. So let's plug in all the numbers and then get r2 out of the way. First thing, however, I want to show you is that once I draw the r1 vectors, this is r1, the vector. Remember, it's an arrow from the axle rotation to the point where the force happens. So that's r1 and then this is r2. Notice that the mg's will make an angle of 90 degrees with the

A 20 kg, 5 m-long bar of uniform mass distribution is attached to the ceiling by a light string, as shown. Because the string is off-center (2 m from the right edge), the bar does not hang horizontally. To fix this, you place a small object on the right edge of the bar. What mass should this object have, to cause the bar to balance horizontally?

• A. 2 kg
• B. 4 kg
• C. 5 kg
• D. 10 kg

Two kids (m,_LEFT = 50 kg, m,_RIGHT = 40 kg) sit on the very ends of a 5 m-long, 30 kg seesaw. How far from the left end of the seesaw should the fulcrum be placed so the system is at equilibrium? (Remember the weight of the seesaw!)

• A. 2.29 m
• B. 2.50 m
• C. 2.71 m
• D. 4.58 m

Multiple objects hanging

Hey guys. In this example, we have a weird arrangement of objects that are sort of hanging together by a bunch of ropes and rods, and we want to find the missing masses. So we don't know the mass of A and C. We're going to call this guy B and we're going to say the mass of B is 4. So we want to find the masses of A and C. And we also want to know all these tensions. Right? We want to know all the tensions. So calculate the 5 vertical, the tension of the 5 ropes, and the 2 missing masses. It says that the system is in linear and rotational equilibrium. This means the sum of all forces equals 0. It also means that the sum of all torques equals 0. It also tells us that all the ropes, all the vertical ropes, which are the arrows are massless, and all the horizontal rods, which are the horizontal lines, are massless as well. So we don't have to worry about those. And we're going to use g as 10 meters per second squared to simplify things. Alright. So first of all, I want to name a few things here. We have A, B, and C. Now this is going to be the tension of C because it holds C. This is going to be the tension of B because it holds B. This tension here, I'm going to call this the tension of B and C because it holds those two masses. This is going to be TA. Whoops. That doesn't go there. This is going to be TA. And this one holds A, B, and C, so this is going to be TABC. Let's talk about forces first. All the forces cancel. So let's look at each object. This here has a force of M_Bg. And because this object is in equilibrium, the tension up cancels the force down. I can actually calculate M_Bg because I have the mass as 4, gravity is 10, we're using 10. So this whole thing is 40. Therefore, this has to be 40. Here, I can't find, I can't tell what T_C is yet because I don't know M_C. So we're going to have to use something else. If I look at this stock here, you can tell that this stock is being pulled up by T_BC but pulled down by these two guys here. Okay. So T_BC is going to be T_B+T_C. If you look at A, I have M_Ag, but I don't know what that is yet, so I can't find TA just yet. I can also see how T_ABC is T_A plus T_BC. That's why we call it ABC because it's all three of them. Okay? So there's a bunch of stuff we don't know yet. We have to find these two masses which will allow us to find these two tensions. And then once we have that, we're missing this, that's why we can't find this. And then once we have that, we'll be able to find the whole thing. We're also missing all of this stuff up here. Okay. You want to start this problem from the ends of the problem. What I mean is this is like a bunch of masses that are chained together. This is the lower point, the end of the problem. The reason I call the end of the problem is because we're going to get to this point, we're gonna need this point first and A will contribute and then this stuff will contribute as well. So the first thing we have to do is work our way from the bottom here, up. Right? And then once we get to this point here, we'll be able to figure out what A is. Alright. So let's just get started. So, everything's figured out for B. So we're gonna go with C. We can't figure out C using F, using the sum of all forces equals 0 because we already did that. All that tells us is that T_C equals M_Cg. But I don't know these two guys, so I'm stuck. So what you have to do is you have to use torque. And you're going to have to say we're going to say that the torque about this point here, let's call this point 1, the sum of all torques about point 1 is 0, which means that all the torques acting on it will cancel out. There's two torques acting about that point. First of all, T_B is doing this. So this is torque of T_B and T_C is doing this, torque of T_C. So imagine you're the center point, there's a rope that pulls you down this way and there's a rope that pulls you down this way. Right? And those two torques will cancel. So I can write T_B equals T_C. Now a torque is a force, which in this case is T_B. The distance, which we're going to call R_B, and then sine of theta_B. And I'm going to write the same thing for the right side, T_CR_C sin of theta_C. The r vector would be this here. Right? From the axis, this is the point we're treating as the axis of rotation, to this point here. So this is R_B and this is R_C. Notice that in both cases, the axes are this way, the tensions are down. So in both cases, the angles are 90. So this is going to be a 1 and a 1. So this simplifies to T_BR_B equals to T_CR_C. And by the way, that's going to be the case with every one of these kinds of points. So when you go here to point 2, we're gonna be able to write the same thing. There are no angles here for us to worry about. Alright? So I know that T_B is 40. R_B is the distance. It says right there, the distance is 1. Then I'm going to be able to find T_C because I have the distance here for R_C is 2. So T_C equals 40 divided by 2. T_C is 20. Okay. T_C is 20. So now that I know that T_C is 20, I can plug it in here, 20 equals M_C 10. That means M_C is 2. Okay? So not only we figured out the tension, but we also figured out that M_C will be 2 kilograms. Okay? Now there's a faster way you could have done this once you get good at this stuff. It's basically a little game. Once you realize that this is a relationship, you can sort of say, hey, you know what? This guy is, this guy here is whoops. C is twice the distance. So to have the same torque, you have to have half the mass. Right? Because torque, in this case of the weight is M_GR and here the sine of theta just gives us 1. So if you have if you have double the distance, half the mass. Cool. Now now that we know T_C, we can actually find T_BC. Right? T_BC is T_B, which is 40 +20=60 60 newtons. Okay? 60 newtons. Now check it out. You can think of this tension, this tension right here as essentially holding the 4 and the 2. Right? So you can think of this as A and 6. You can combine those two, the 4 and the 2, into a 6. And then there's a tension here. This is 1 and this is 4. And remember what I just told you, if you are farther, you have to be lighter. A, if it's 4 times as far from the central point here, from the axis of rotation, from the support point than 6. So if it's 4 times farther, it also has to be 4 times lighter. And just by using this, you can tell that the mass of A has to be 6 over 4 which is 1.5. So you can do it like this very quickly. Alright? Now I'm going to do this full way which would be to write, here we wrote that all torques on point 1 are 0. Now I'm going to write that all torques on point 2 are 0. And what this gives us is that the torque of A will cancel out with the torque of BC. Torque of A equals torque of BC. So tension of A R_A sin theta_A equals tension_BCR_BCsin of theta_BC. We just talked about, we talked about earlier how the sines will just be 1, because all the r vectors are horizontal and all the tensions are vertical. The tension we're looking for here is A. The distance is 4 meters. So T_A 4. T_B is 60. The distance for T_B is 1. So it's right here, 1. So T_A is 60 over 4=15 newtons. So we know that this is 15 newtons and these two guys have to be the same. So if T_A is 15, then M_A, gravity is 10. M_A is 10 divided by 15=1.5, which is what I just mentioned here, but I wanted to show you how this stuff works out. Okay. So you can actually skip some steps if you just realize that, you have this relationship here at any one of these points that connect stuff. Right? You can write you can actually also write this like this. M_BR_B equals T_C I'm sorry, M_CR_C, which is essentially what I did when I shortcutted this question. The reason you can do this is because the torque equation does this, M_BgR_B equals M_CgR_C, and then you're essentially canceling the r's and ending up with something simpler like this. Okay. So that's a quick shortcut you can use. I almost have everything. The last thing I have to do is I have to combine the 60 here with the 15 here, and I get a 75 newtons as the tension of everything. So I got the masses. The masses are 1.5, 4, and 2. The tensions are 75, 15, 60, 40, and 20. K? And all it took was using the fact that the tension's up equals the masses of the weights down, and the fact that you can use torque about points 1 2. The fact that the torques cancel, and you end up with stuff like this and stuff like this that allows you to find the other ones. Okay? So it's a very peculiar kind of question, but once you get the hang of this, it's really easy to solve a bunch of them. So hopefully it made sense. Let me know if you guys have any questions and let's keep going.